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I am trying to solve the following nonlinear system of equations. Could someone kindly give me some tips on how I can solve it ?

$$ \left \{ \begin{array}{c} l^2= (x_3-x_4 )^2+(y_3-y_4 )^2+(z_3-z_4 )^2 (1) \\ (d_1+a_1)^2= (x_3-x_1 )^2+(y_3-y_1 )^2+(z_3-z_1 )^2 (2)\\ (d_2+a_2)^2= (x_3-x_2 )^2+(y_3-y_2 )^2+(z_3-z_2 )^2 (3)\\ (d_3+a_3)^2= (x_4-x_1 )^2+(y_4-y_1 )^2+(z_4-z_1 )^2 (4)\\ (d_4+a_4)^2= (x_4-x_2 )^2+(y_4-y_2 )^2+(z_4-z_2 )^2 (5)\\ z_4= z_3+l\sin(θ) (6)\\ \end{array} \right. $$ There are four points.The coordinates of point$1(x_1,y_1,z_1)$ and point$2(x_2,y_2,z_2)$ are known. I want to calculate the coordinates of point$3(x_3,y_3,z_3)$ and point$4(x_4,y_4,z_4)$.The distance between point$3(x_3,y_3,z_3)$ and point$4(x_4,y_4,z_4)$ is $l$ which has been known. $θ$ has also been known which means the inclination angle between point$3$ and point$4$. $d_1,d_2,d_3,d_4$ means the distances between points. However the distance $d_1,d_2,d_3,d_4$ are estimated value，the estimated errors are $a_1,a_2,a_3,a_4$ .The values of $d_1+a_1,d_2+a_2,d_3+a_3,d_4+a_4$ have been known. Based on the conditions above, how to get the coordinates of point $3(x_3,y_3,z_3)$ and point $4(x_4,y_4,z_4)$.

For Newton's Method,if i don't have the estimated errors a1,a2,a3,a4, exact solution may be obtained.
$$ \left \{ \begin{array}{c} l^2= (x_3-x_4 )^2+(y_3-y_4 )^2+(z_3-z_4 )^2 (1) \\ (d_1)^2= (x_3-x_1 )^2+(y_3-y_1 )^2+(z_3-z_1 )^2 (2)\\ (d_2)^2= (x_3-x_2 )^2+(y_3-y_2 )^2+(z_3-z_2 )^2 (3)\\ (d_3)^2= (x_4-x_1 )^2+(y_4-y_1 )^2+(z_4-z_1 )^2 (4)\\ (d_4)^2= (x_4-x_2 )^2+(y_4-y_2 )^2+(z_4-z_2 )^2 (5)\\ z_4= z_3+l\sin(θ) (6)\\ \end{array} \right. $$ But when i have the estimated errors a1,a2,a3,a4,Will Newton's iteration work?I can't get the value of d1,d2,d3,d4,a1,a2,a3,a4,i can get the value of (d1+a1),(d2+a2),(d3+a3),(d4+a4) $$ \left \{ \begin{array}{c} l^2= (x_3-x_4 )^2+(y_3-y_4 )^2+(z_3-z_4 )^2 (1) \\ (d_1+a_1)^2= (x_3-x_1 )^2+(y_3-y_1 )^2+(z_3-z_1 )^2 (2)\\ (d_2+a_2)^2= (x_3-x_2 )^2+(y_3-y_2 )^2+(z_3-z_2 )^2 (3)\\ (d_3+a_3)^2= (x_4-x_1 )^2+(y_4-y_1 )^2+(z_4-z_1 )^2 (4)\\ (d_4+a_4)^2= (x_4-x_2 )^2+(y_4-y_2 )^2+(z_4-z_2 )^2 (5)\\ z_4= z_3+l\sin(θ) (6)\\ \end{array} \right. $$ As we konw,by equation(2) minus equation(3),we can obtain a plane P1. The point3 is located on intersection circle between the plane P1 and sphere Point1.Similarly,by equation(4) minus equation(5),the point4 is located on another intersection circle.what i want to ask is that how to use the equation(1) and equation(6) to construct a cost function?If i traversal coordinates on intersection circles,will i get the optimal solution?

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  • $\begingroup$ If $(d_i+a_i)$ are known then your equations have no estimated values in them? Newton's Method in several dimensions should solve this system. The geometry of the problem (some points on intersecting spheres maybe?) should indicate a good initial approximate solution for the method. $\endgroup$ Commented Dec 6, 2021 at 14:55
  • $\begingroup$ Can you consider optimization ? If yes, I shall elaborate. $\endgroup$ Commented Dec 6, 2021 at 15:11
  • $\begingroup$ I have tried Newton's Method,but the coordinates of point3 and point4 are in a large range.I can not give a reasonable starting value,so the calculation results are not satisfactory.The geometry of the problem may help to solve the problem,i need time to think about it. $\endgroup$ Commented Dec 7, 2021 at 2:02
  • $\begingroup$ This is the problem about engineering,so i just need to get optimal solution.Because the symmetry of equations,may be there are multiple solutions. i have no idea about how to get optimal solution. $\endgroup$ Commented Dec 7, 2021 at 2:17
  • $\begingroup$ @ Claude Leibovici yes,please,i want to get the optimal solution $\endgroup$ Commented Dec 9, 2021 at 5:38

1 Answer 1

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You set $\mathbf{x}_0$ to some reasonable starting value, for example to \begin{align} \mathbf{x}_0&=\left[\begin{array}{c}x_1\\y_1\\z_1\\x_2\\y_2\\z_2\end{array}\right] \end{align} and then you iterate $\mathbf{x}_{i}=-\left[\frac{\partial\mathbf{F}}{\partial\mathbf{x}}(\mathbf{x}_{i-1})\right]^{-1}\mathbf{F}(\mathbf{x}_{i-1}),~i=1,2,\dots$ until convergence, where \begin{align} \mathbf{x}&=\left[\begin{array}{c}x_3\\y_3\\z_3\\x_4\\y_4\\z_4\end{array}\right],~ \mathbf{F}(\mathbf{x})=\left[\begin{array}{c} (x_3-x_4)^2+(y_3-y_4)^2+(z_3-z_4)^2-l^2\\ (x_3-x_1)^2+(y_3-y_1)^2+(z_3-z_1)^2-(d_1+a_1)^2\\ (x_3-x_2)^2+(y_3-y_2)^2+(z_3-z_2)^2-(d_2+a_2)^2\\ (x_4-x_1)^2+(y_4-y_1)^2+(z_4-z_1)^2-(d_3+a_3)^2\\ (x_4-x_2)^2+(y_4-y_2)^2+(z_4-z_2)^2-(d_4+a_4)^2\\ z_3-z_4+l\sin\theta \end{array}\right],\\ \frac{\partial\mathbf{F}(\mathbf{x})}{\partial\mathbf{x}}&=\left[\begin{array}{cccccc} 2(x_3-x_4)&2(y_3-y_4)&2(z_3-z_4)& -2(x_3-x_4)&-2(y_3-y_4)&-2(z_3-z_4)\\ 2(x_3-x_1)&2(y_3-y_1)&2(z_3-z_1)&0&0&0\\ 2(x_3-x_2)&2(y_3-y_2)&2(z_3-z_2)&0&0&0\\ 0&0&0&2(x_4-x_1)&2(y_4-y_1)&2(z_4-z_1)\\ 0&0&0&2(x_4-x_2)&2(y_4-y_2)&2(z_4-z_2)\\ 0&0&1&0&0&-1\\ \end{array}\right]. \end{align}

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  • $\begingroup$ The question is i can not give a reasonable starting value for point3 and point4 moving in space.Because the symmetry of equations,may be there are multiple solutions. $\endgroup$ Commented Dec 7, 2021 at 2:08
  • $\begingroup$ You have to try different starting values and show what makes the Newton method to converge. The Newton method finds only one of all local solutions (and only in case of convergece). If you need to find all existing solutions, you can use the multistart approach for example, in which you run the method let say $10^5$ times with always different, randomly generated starting points. I expect to have some box-constrain in which the points are expected. $\endgroup$ Commented Dec 7, 2021 at 7:52
  • $\begingroup$ You can also reduce dimensionality of the problem to 5 by inserting the last equation to other ones. $\endgroup$ Commented Dec 7, 2021 at 7:56

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